Integration Methods - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Methods of Integration-1”.

1. Find ∫7 cos⁡mx dx.
a) \(\frac{7 \,sin⁡mx}{x}+C\)
b) \(\frac{7 \,sin⁡mx}{m}+C\)
c) \(\frac{sin⁡mx}{x}+C\)
d) \(\frac{sin⁡x}{m}+C\)
View Answer

Answer: b
Explanation: Using Integration by Substitution, Let xm=t
Differentiating w.r.t x, we get
mdx=dt
∴\(\int 7 \,cos⁡mx \,dx=\int \frac{(7 cos⁡t)}{m} dt\)
=\(\frac{7}{m} \int cos⁡t \,dt=\frac{7}{m} (sin⁡t)+C\)
Replacing t with mx again we get,
\(\int 7 \,cos⁡mx \,dx=\frac{7 \,sin⁡mx}{m}+C\)

2. Integrate \(3x^2 (cos⁡x^3+8)\).
a) \(sin⁡x^3-8x^3+C\)
b) \(sin⁡x^3+8x^3+C\)
c) –\(sin⁡x^3+8x^3+C\)
d) \(sin⁡x^3-x^3+C\)
View Answer

Answer: b
Explanation: By using the method of integration by substitution,
Let x³=t
Differentiating w.r.t x, we get
3x² dx=dt
\(\int 3x^2 \,(cos⁡x^3+8) \,dx=\int (cos⁡t+8)dt\)
\(\int (cos⁡t+8) dt=sin⁡t+8t\)
Replacing t with x³,we get
\(\int 3x^2 (cos⁡x^3+8) dx=sin⁡x^3+8x^3+C\)

3. Find \(\int 6x(x^2+6)dx\).
a) \(\frac{3x^4}{2}+18x^2+C\)
b) \(\frac{3x^4}{2}-18x+C\)
c) \(\frac{3x^4}{2}-18x^2+C\)
d) \(\frac{3x^4}{2}+x^2+C\)
View Answer

Answer: a
Explanation: Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
\(\int 6x(x^2+6)dx=3\int (t+6) dt\)
3\(\int (t+6)dt=3\left (\frac{t^2}{2}+6t\right )=\frac{3t^2}{2}+18t\)
Replacing t with x²
\(\int 6x(x^2+6)dx=\frac{3x^4}{2}+18x^2+C\)

4. Find the integral of \(3e^x+\frac{2(log⁡ x)}{3x}\).
a) \(3e^x+\frac{1}{3} (x)^2+C\)
b) \(e^x-\frac{8}{3} (log⁡x)^2+C\)
c) \(3e^x-\frac{1}{3} (log⁡x)^2+C\)
d) \(3e^x+\frac{1}{3} (log⁡x)^2+C\)
View Answer

Answer: d
Explanation: \(\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3\int e^x dx+\frac{2}{3} \int \frac{log⁡x}{x}\)
Let log⁡x=t
Differentiating w.r.t x, we get
\(\frac{1}{x} dx=dt\)
∴\(\int \frac{log⁡x}{x}=\int \,t \,dt=\frac{t^2}{2}\)
\(\int e^x dx=e^x\)
Replacing t with log⁡x, we get
\(\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3e^x+\frac{1}{3} (log⁡x)^2+C\)

5. Find \(\int \frac{e^{-cot^{-1}⁡x}}{1+x^2}\).
a) \(e^{-cot^{-1}⁡x}+C\)
b) \(e^{-2cot^{-1}⁡x}+C\)
c) \(e^{-tan^{-1}⁡x}+C\)
d) \(e^{-cot^1⁡2x}+C\)
View Answer

Answer: a
Explanation: Let \(-cot^{-1}⁡x\)=t
Differentiating w.r.t x, we get
–\(\left (-\frac{1}{1+x^2}\right )dx=dt\)
\(\frac{1}{1+x^2} dx=dt\)
\(\int \frac{e^{-cot^{-1}}x}{1+x^2} dx=\int e^t \,dt\)
=e^t
Replacing t with -cot^-1x, we get
\(\int \frac{e^{-cot^{-1}}x}{1+x^2} dx=e^{-cot^{-1}}x+C\)

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6. Find the integral of \(\frac{5x^4}{\sqrt{x^5+9}}\).
a) \(\sqrt{x^5+9}\)
b) \(2\sqrt{x^5-9}\)
c) 2(x⁵+9)
d) \(2\sqrt{x^5+9}\)
View Answer

Answer: d
Explanation: Let x⁵+9=t
Differentiating w.r.t x, we get
5x⁴ dx=dt
\(\int \frac{5x^4}{\sqrt{x^5+9}} dx=\int \frac{dt}{\sqrt{t}}\)
=\(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2\sqrt{t}\)
Replacing t with x⁵+9, we get
\(\int \frac{5x^4}{\sqrt{x^5+9}} dx=2\sqrt{x^5+9}\).

7. Find \(\int \frac{6 sin⁡\sqrt{x}}{\sqrt{x}} dx\)
a) \(2 \,cos⁡\sqrt{x}+C\)
b) –\(12 \,cos⁡\sqrt{x}+C\)
c) -12 cos⁡x+C
d) 12 cos⁡x+C
View Answer

Answer: b
Explanation: Let \(\sqrt{x}=t\)
Differentiating w.r.t x,we get
\(\frac{1}{2\sqrt{x}} dx=dt\)
\(\frac{1}{\sqrt{x}} dx=2dt\)
∴\(\int \frac{6 sin⁡\sqrt{x}}{\sqrt{x}} dx=\int \,12 \,sin⁡t \,dt\)
=12(-cos⁡t)=-12 cos⁡t
Replacing t with \(\sqrt{x}\), we get
\(\int \frac{6 sin⁡\sqrt{x}}{\sqrt{x}} dx=-12 \,cos⁡\sqrt{x}+C\)

8. Find \(\int \frac{20x^3}{1+x^4} dx\).
a) 5 log⁡(x⁴)+C
b) -5 log⁡(1+x⁴)+C
c) 5 log⁡(1+x⁴)+C
d) log⁡(1+x⁴)+C
View Answer

Answer: c
Explanation: Let 1+x⁴=t
4x³ dx=dt
∴\(\int \frac{20x^3}{1+x^4} dx=5\int \frac{dt}{t}\)
=5 log⁡t
Replacing t with 1+x⁴, we get
\(\int \frac{20x^3}{1+x^4} dx=5 \,log⁡(1+x^4)+C\)

9. Integrate \(\frac{x^2}{e^{x^3}}\).
a) –\(\frac{1}{(3e^{x^3})}+C\)
b) \(\frac{1}{3e^{x^3}}+C\)
c) –\(\frac{1}{e^{x^3}}+C\)
d) e^x³+C
View Answer

Answer: a
Explanation: Let x³=t
3x² dx=dt
x² dx=dt/3
∴\(\int \frac{x^2}{e^{x^3}} dx=\frac{1}{3} \int \frac{dt}{e^t}\)
=\(\frac{1}{3} \left (-e^{-t}\right )\)
Replacing t with x³, we get
\(\int \frac{x^2}{e^{x^3}} dx=-\frac{1}{3e^{x^3}}+C\)

10. Find \(\int \frac{cos^{-1}x}{\sqrt{1-x^2}} dx\).
a) \(\frac{(sin^{-1}⁡x)^2}{2}+C\)
b) \(\frac{(cos^{-1}⁡x)^2}{7}+C\)
c) \(\frac{(cos^{-1}⁡x)^2}{2}+C\)
d) –\(\frac{(cos^{-1}⁡x)^2}{2}+C\)
View Answer

Answer: c
Explanation: Let cos^-1⁡x=t
Differentiating w.r.t x, we get
\(\frac{1}{\sqrt{1-x^2}} dx=dt\)
∴\(\int \frac{cos^{-1}⁡x}{\sqrt{1-x^2}} dx=\int t dt\)
=\(\frac{t^2}{2}\)
Replacing t with cos^-1x,we get
\(\int \frac{cos^{-1}⁡x}{\sqrt{1-x^2}} dx=\frac{(cos^{-1}⁡x)^2}{2}+C\)

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